ScienceAges 16-18

Acid-base titration stoichiometry

Use balanced diprotic neutralisation stoichiometry (H₂SO₄ + 2NaOH) to calculate an unknown alkali concentration from titration volumes. Review suitability for your learners before teaching.

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Worksheet StudioPractice pack · Ages 16-18

Acid-base titration stoichiometry

Take your time. Show your thinking.

1. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 0.8 M · 2NaOH: 40 cm³ × ? M
2. In an acid-base titration, 20 cm³ of 3.2 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 3.2 M · 2NaOH: 40 cm³ × ? M
3. In an acid-base titration, 25 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 4 M · 2NaOH: 40 cm³ × ? M
4. In an acid-base titration, 50 cm³ of 3.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 3.5 M · 2NaOH: 50 cm³ × ? M
5. In an acid-base titration, 30 cm³ of 0.25 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 0.25 M · 2NaOH: 25 cm³ × ? M
6. In an acid-base titration, 30 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 0.5 M · 2NaOH: 25 cm³ × ? M
7. In an acid-base titration, 10 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 10 cm³ × 0.5 M · 2NaOH: 10 cm³ × ? M
8. In an acid-base titration, 40 cm³ of 0.4 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 0.4 M · 2NaOH: 10 cm³ × ? M
9. In an acid-base titration, 30 cm³ of 1.1 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 1.1 M · 2NaOH: 10 cm³ × ? M
10. In an acid-base titration, 40 cm³ of 2 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 2 M · 2NaOH: 25 cm³ × ? M
11. In an acid-base titration, 30 cm³ of 1.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 1.5 M · 2NaOH: 50 cm³ × ? M
12. In an acid-base titration, 40 cm³ of 0.7 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 0.7 M · 2NaOH: 10 cm³ × ? M

12 questions · Seed 104729 · Age guidance, not curriculum certification

View the answer key

Worksheet StudioAnswer key · Ages 16-18

Acid-base titration stoichiometry

Answers and reasoning. Review before teaching.

1. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 50) ÷ 40 = 2 mol/dm³.
2. In an acid-base titration, 20 cm³ of 3.2 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.2 × 20) ÷ 40 = 3.2 mol/dm³.
3. In an acid-base titration, 25 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
5 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4 × 25) ÷ 40 = 5 mol/dm³.
4. In an acid-base titration, 50 cm³ of 3.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
7 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.5 × 50) ÷ 50 = 7 mol/dm³.
5. In an acid-base titration, 30 cm³ of 0.25 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
0.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.25 × 30) ÷ 25 = 0.6 mol/dm³.
6. In an acid-base titration, 30 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.5 × 30) ÷ 25 = 1.2 mol/dm³.
7. In an acid-base titration, 10 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.5 × 10) ÷ 10 = 1 mol/dm³.
8. In an acid-base titration, 40 cm³ of 0.4 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.4 × 40) ÷ 10 = 3.2 mol/dm³.
9. In an acid-base titration, 30 cm³ of 1.1 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
6.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.1 × 30) ÷ 10 = 6.6 mol/dm³.
10. In an acid-base titration, 40 cm³ of 2 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
6.4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 2 × 40) ÷ 25 = 6.4 mol/dm³.
11. In an acid-base titration, 30 cm³ of 1.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.8 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.5 × 30) ÷ 50 = 1.8 mol/dm³.
12. In an acid-base titration, 40 cm³ of 0.7 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
5.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.7 × 40) ÷ 10 = 5.6 mol/dm³.

12 questions · Seed 104729 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 16-18

Acid-base titration stoichiometry

Take your time. Show your thinking.

1. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 0.8 M · 2NaOH: 40 cm³ × ? M
2. In an acid-base titration, 20 cm³ of 3.2 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 20 cm³ × 3.2 M · 2NaOH: 40 cm³ × ? M
3. In an acid-base titration, 25 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 25 cm³ × 4 M · 2NaOH: 40 cm³ × ? M
4. In an acid-base titration, 50 cm³ of 3.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 50 cm³ × 3.5 M · 2NaOH: 50 cm³ × ? M
5. In an acid-base titration, 30 cm³ of 0.25 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 0.25 M · 2NaOH: 25 cm³ × ? M
6. In an acid-base titration, 30 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 0.5 M · 2NaOH: 25 cm³ × ? M
7. In an acid-base titration, 10 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 10 cm³ × 0.5 M · 2NaOH: 10 cm³ × ? M
8. In an acid-base titration, 40 cm³ of 0.4 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 0.4 M · 2NaOH: 10 cm³ × ? M
9. In an acid-base titration, 30 cm³ of 1.1 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 1.1 M · 2NaOH: 10 cm³ × ? M
10. In an acid-base titration, 40 cm³ of 2 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 2 M · 2NaOH: 25 cm³ × ? M
11. In an acid-base titration, 30 cm³ of 1.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 30 cm³ × 1.5 M · 2NaOH: 50 cm³ × ? M
12. In an acid-base titration, 40 cm³ of 0.7 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
H₂SO₄: 40 cm³ × 0.7 M · 2NaOH: 10 cm³ × ? M

12 questions · Seed 104729 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 16-18

Acid-base titration stoichiometry

Answers and reasoning. Review before teaching.

1. In an acid-base titration, 50 cm³ of 0.8 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.8 × 50) ÷ 40 = 2 mol/dm³.
2. In an acid-base titration, 20 cm³ of 3.2 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.2 × 20) ÷ 40 = 3.2 mol/dm³.
3. In an acid-base titration, 25 cm³ of 4 mol/dm³ H₂SO₄ completely neutralises 40 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
5 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 4 × 25) ÷ 40 = 5 mol/dm³.
4. In an acid-base titration, 50 cm³ of 3.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
7 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 3.5 × 50) ÷ 50 = 7 mol/dm³.
5. In an acid-base titration, 30 cm³ of 0.25 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
0.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.25 × 30) ÷ 25 = 0.6 mol/dm³.
6. In an acid-base titration, 30 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.5 × 30) ÷ 25 = 1.2 mol/dm³.
7. In an acid-base titration, 10 cm³ of 0.5 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.5 × 10) ÷ 10 = 1 mol/dm³.
8. In an acid-base titration, 40 cm³ of 0.4 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
3.2 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.4 × 40) ÷ 10 = 3.2 mol/dm³.
9. In an acid-base titration, 30 cm³ of 1.1 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
6.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.1 × 30) ÷ 10 = 6.6 mol/dm³.
10. In an acid-base titration, 40 cm³ of 2 mol/dm³ H₂SO₄ completely neutralises 25 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
6.4 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 2 × 40) ÷ 25 = 6.4 mol/dm³.
11. In an acid-base titration, 30 cm³ of 1.5 mol/dm³ H₂SO₄ completely neutralises 50 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
1.8 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 1.5 × 30) ÷ 50 = 1.8 mol/dm³.
12. In an acid-base titration, 40 cm³ of 0.7 mol/dm³ H₂SO₄ completely neutralises 10 cm³ of NaOH solution according to H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the NaOH solution in mol/dm³.
5.6 mol/dm³
By the 1:2 stoichiometry, 2 × c(H₂SO₄) × V(H₂SO₄) = c(NaOH) × V(NaOH), so c(NaOH) = (2 × 0.7 × 40) ÷ 10 = 5.6 mol/dm³.

12 questions · Seed 104729 · Age guidance, not curriculum certification